Download A characterization of Symmetric group Sr, where r is prime number

Annales Mathematicae et Informaticae
40 (2012) pp. 13–23
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A characterization of Symmetric group Sr ,
where r is prime number
Alireza Khalili Asboeia , Seyed Sadegh Salehi Amiria
Ali Iranmaneshb , Abolfazl Tehraniana
a
Department of Mathematics, Science and Research Branch
Islamic Azad University, Tehran, Iran
[email protected], [email protected], [email protected]
b
Department of Mathematics, Faculty of Mathematical Sciences
Tarbiat Modares University, Tehran, Iran
[email protected]
Submitted November 9, 2011 — Accepted April 11, 2012
Abstract
Let G be a finite group and πe (G) be the set of element orders of G.
Let k ∈ πe (G) and mk be the number of elements of order k in G. Set
nse(G) := {mk | k ∈ πe (G)}. In this paper, we prove the following results:
1. If G is a group such that nse(G) = nse(Sr ), where r is prime number
and |G| = |Sr |, then G ∼
= Sr .
2. If G is a group such that nse(G) = nse(Sr ), where r < 5 × 108 and r − 2
are prime numbers and r is a prime divisor of |G|, then G ∼
= Sr .
Keywords: Element order, set of the numbers of elements of the same order,
Symmetric group
MSC: 20D06, 20D20, 20D60
1. Introduction
If n is an integer, then we denote by π(n) the set of all prime divisors of n. Let
G be a finite group. Denote by π(G) the set of primes p such that G contains an
element of order p. Also the set of element orders of G is denoted by πe (G). A
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A. K. Asboei, S. S. S. Amiri, A. Iranmanesh, A. Tehranian
finite group G is called a simple Kn -group, if G is a simple group with |π(G)| = n.
Set mi = mi (G) := |{g ∈ G| the order of g is i}| and nse(G) := {mi |i ∈ πe (G)}.
In fact, mi is the number of elements of order i in G and nse(G) is the set of
sizes of elements with the same order in G. Throughout this paper, we denote
by φ the Euler’s totient function. If G is a finite group, then we denote by Pq
a Sylow q-subgroup of G and by nq (G) the number of Sylow q-subgroup of G,
that is, nq (G) = |Sylq (G)|. Also we say pk k m if pk | m and pk+1 - m. For a
real number x, let ϕ(x) denote the number of primes which are not greater than
x, and [x] the greatest integer not exceeding x. For positive integers n and k,
Qk Q
let tn (k) = i=1 ( n/(i+1)<p≤n/i p)i , where p is a prime. Denote by gcd(a, b) the
greatest common divisor of positive integers a and b, and by expm (a) the exponent
of a modulo m for the relatively prime integers a and m with m > 1. If m is a
positive integer and p is a prime, let |m|p denote the p-part of m; in the other
words, |m|p = pk if pk | m but pk+1 - m. For a finite group H, |H|p denotes the
p-part of |H|. All further unexplained notations are standard and refer to [1], for
example. In [2] and [3], it is proved that all simple K4 -groups and Mathieu groups
can be uniquely determined by nse(G) and the order of G. In [4], it is proved
that the groups A4 , A5 and A6 are uniquely determined only by nse(G). In [5],
the authors show that the simple group P SL(2, q) is characterizable by nse(G) for
each prime power 4 ≤ q ≤ 13. In this work it is proved that the Symmetric group
Sr , where r is a prime number is characterizable by nse(G) and the order of G. In
fact the main theorems of our paper are as follow:
Theorem 1. Let G be a group such that nse(G)=nse(Sr ), where r is a prime
number and |G| = |Sr |. Then G ∼
= Sr .
Theorem 2. Let G be a group such that nse(G)=nse(Sr ), where r < 5 × 108 and
r − 2 are prime numbers and r ∈ π(G). Then G ∼
= Sr .
In this paper, we use from [6] for proof some Lemmas, but since some part of
the proof is different, we were forced to prove details get’em. We note that there
are finite groups which are not characterizable by nse(G) and |G|. For example see
the Remark in [2].
2. Preliminary Results
We first quote some lemmas that are used in deducing the main theorems of this
paper.
Let α ∈ Sn be a permutation and let α have ti cycles of length i, i = 1, 2, . . . , l,
in its cycle decomposition. The cycle structure of α is denote by 1t1 2t2 . . . ltl ,
where 1t1 + 2t2 · · · + ltl = n. One can easily show that two permutations in Sn are
conjugate if and only if they have the same cycle structure.
Lemma 2.1 ([6]).
(i) ϕ(x) − ϕ(x/2) ≥ 7 for x ≥ 59.
A characterization of Symmetric group Sr , where r is prime number
15
(ii) ϕ(x) − ϕ(x/4) ≥ 12 for x ≥ 61.
(iii) ϕ(x) − ϕ(6x/7) ≥ 1 for x ≥ 37.
Lemma 2.2 ([6]). If n ≥ 402, then (2/n)tn (6) > e1.201n . If n ≥ 83, then
(2/n)tn (6) > e0.775n .
Lemma 2.3 ([6]). Let p be a prime and k a positive integer.
(i) If |n!|p = pk , then (n − 1)/(p − 1) ≥ k ≥ n/(p − 1) − 1 − [logp n].
(ii) If |n!/m!|p = pk and 0 ≤ m < n, then k ≤ (n − m − 1)/(p − 1) + [logp n].
Lemma 2.4 ([7]). Let α ∈ Sn and assume that the cycle decomposition of α
contains t1 cycles of length 1, t2 cycles of length 2, . . . , tl cycles of length l. Then
the order of conjugacy class of α in Sn is n!/1t1 2t2 . . . ltl t1 !t2 ! . . . tl !.
Lemma 2.5 ([8]). Let G be a finite group and m be a positive integer dividing |G|.
If Lm (G) = {g ∈ G|g m = 1}, then m | |Lm (G)|.
Lemma 2.6 ([9]). Let G be a finite group and p ∈ π(G) be odd. Suppose that P
is a Sylow p-subgroup of G and n = ps m, where (p, m) = 1. If P is not cyclic and
s > 1, then the number of elements of order n in G is always a multiple of ps .
Lemma 2.7 ([4]). Let G be a group containing more than two elements. Let
k ∈ πe (G) and mk be the number of elements of order k in G. If s = sup{mk |k ∈
πe (G)} is finite, then G is finite and |G| ≤ s(s2 − 1).
Let mn be the number of elements of order n. We note that mn = kφ(n), where
k is the number of cyclic subgroups of order n in G. Also we note that if n > 2,
then φ(n) is even. If n ∈ πe (G), then by Lemma 2.2 and the above notation we
have
(
φ(n) | mn
P
(2.1)
n | d|n md
In the proof of the main theorem, we often apply (2.1) and the above comments.
3. Proof of the Main Theorem 1
We now prove the theorem 1 stated in the introduction. Let G be a group such
that nse(G) = nse(Sr ), where r is a prime number and |G| = |Sr |. The following
Lemmas reduce the problem to a study of groups with the same order with Sr .
Lemma 3.1. mr (G) = mr (Sr ) = (r − 1)! and if S ∈ Sylr (G), R ∈ Sylr (Sr ), then
|NG (S)| = |NSr (R)|.
Proof. Since mr (G) ∈ nse(G) and nse(G) = nse(Sr ), then by (2.1)
P there exists
k ∈ πe (Sr ) such that p | 1 + mk (Sr ). We know that mk (Sr ) =
|clSr (xi )| such
that |xi | = k. Since r | 1 + mk (Sr ), then (r, mk (Sr )) = 1. If the cyclic structure
of xi for any i is 1t1 2t2 . . . ltl such that t1 , t2 , . . . , tl and 1, 2, . . . , l are not equal
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A. K. Asboei, S. S. S. Amiri, A. Iranmanesh, A. Tehranian
to r, then r | r!/1t1 2t2 · · · ltl t1 !t2 ! . . . tl !, that is r | |clSr (xi )| for any i. Therefore
(r, mk (Sr )) 6= 1, which is a contradiction. Thus there exist i ∈ N such that ti = r
or one of the numbers 1, 2, . . . or l is equal to r. If there exist i ∈ N such that
ti = r, then the cyclic structure of xi is 1r . Hence |xi | = 1, which is a contradiction.
If one of the numbers 1, 2, . . . or l is equal to r, then the cyclic structure of xi is
r1 . Hence |xi | = r and k = r. Therefore mr (G) = mr (Sr ), since |G| = |Sr |, then
nr (G) = nr (Sr ) = mr (G)/(r − 1) = (r − 2)!. Hence if S ∈ Sylr (G), R ∈ Sylr (Sr ),
then |NG (S)| = |NSr (R)| = r(r − 1).
Lemma 3.2. G has a normal series 1 ≤ N < H ≤ G such that r | |H/N | and
H/N is a minimal normal subgroup of G/N .
Proof. Suppose 1 = N0 < N1 < · · · < Nm = G is a chief series of G. Then there
exists i such that p | |Ni /Ni−1 |. Let H = Ni and N = Ni−1 . Then 1 ≤ N < H ≤ G
is a normal series of G, H/N is a minimal normal subgroup of G/N , and r | |H/N |.
Clearly, H/N is a simple group.
Lemma 3.3. Let r ≥ 5 and let 1 ≤ N < H ≤ G be a normal series of G, where
H/N is a simple group and r | |H/N |. Let R ∈ Sylr (G) and Q ∈ Sylr (G/N ).
(i) |NG/N (Q)| = |NH/N (Q)||G/H| and |NN (R)||NG/N (Q)| = |NG (R)| = r(r − 1).
(ii) If P ∈ Sylp (N ) with |P | = pk , where p is a prime and k ≥ 1, then either
Qk−1
|H/N | | i=0 (pk − pi ) or pk |NG/N (Q)| | r(r − 1).
Proof. (i) By Frattini’s argument, G/N = NG/N (Q)(H/N ). Thus
G/H ∼
= NG/N (Q)/NH/N (Q).
So the first equality holds. Since we have
NG/N (Q) ∼
= NG (R)N/N ∼
= NG (R)/NN (R),
the second equality is also true.
(ii) By Frattini’s argument again, H = NH (P )N . Thus, we have H/N ∼
=
NH (P )/NN (P ). Since H/N is a simple group, CH (P )NN (P ) = NH (P ) or NN (P ).
If CH (P )NN (P ) = NH (P ), then r | |CH (P )|. Without loss of generality, we may
assume R ≤ CH (P ). It means that NN (R) ≥ P . Then pk |NG/N (Q)| | r(r − 1) by
(i). If CH (P )NN (P ) = NN (P ), then CH (P ) ≤ NN (P ). Thus |NH (P )/NN (P )| |
|NH (P )/CH (P )|. Since |H/N | = |NH (P )/NN (P )| and NH (P )/CH (P ) is isomorQk−1
phic to a subgroup of Aut(P ), |H/N | | |Aut(P )|. Since |Aut(P )| | i=0 (pk − pi ),
Qk−1 k
|H/N | | i=0 (p − pi ).
Lemma 3.4. Let r ≥ 5 and let 1 ≤ N < H ≤ G be a normal series of G with
H/N simple and r | |H/N |. If |N |p |G/H|p = pk with k ≥ 1 and |H/N | not dividing
k
i
k
Πk−1
i=0 (p − p ), then p | (r − 1).
A characterization of Symmetric group Sr , where r is prime number
17
Proof. Assume |N |p = pk . If t = 0, then pk | |G/H|. By Lemma 3.3 (i), pk |
Qk−1
k
i
k
r(r − 1). If t ≥ 1, since |H/N | does not divide Πk−1
i=0 (p − p ) and
i=k−t (p −
Q
Q
t−1
t−1
pi ) = pt(k−t) j=0 (pt − pj ), we have that |H/N | does not divide j=0 (pt − pj ).
By Lemma 3.3 (ii), pt |NG/N (Q)| | r(r − 1), where Q ∈ Sylr (G/N ). By Lemma
3.3 (i), |NG/N (Q)| = |NH/N (Q)||G/H|, so we have pt |G/H| | r(r − 1). Since
|N |p |G/H|p = pk and |N |p = pk , we obtain |G/H|p = pk−t . Thus pk | r(r − 1).
Since r | |H/N |, it is easy to know p 6= r. Therefore, pk | (r − 1).
Lemma 3.5. Let r ≥ 5 and let 1 ≤ N < H ≤ G be a normal series of G with
H/N simple. If r | |H/N |, then tr (1) | |H/N | and H/N is a non-ablian simple
group and G is not solvable group.
Proof. We first prove that tr (1) | |H/N |. If tr (1) - |H/N |, then there exists a prime
p satisfying r/2 < p < r such that p | |N ||G/H|. Since r | |H/N |, |H/N | - (p − 1).
Hence p | (r − 1) by Lemma 3.4. But (r − 1)/2 < r/2, contrary to r/2 < p. Since
the number of prime factors of tr (1) is greater that 1, then H/N is a non-ablian
simple group. Clearly G is not solvable group.
Lemma 3.6. If r ≥ 59 and let 1 ≤ N < H ≤ G be a normal series of G with H/N
simple and r | |H/N |,
(i) If gcd(tr (6), r − 1) = 1, then tr (6) | |H/N |.
(ii) If gcd(tr (6), r − 1) is a prime p, then (tr (6)/p) | |H/N |.
Proof. By Lemma 3.5, tr (1) | |H/N |. Suppose tr (6) - |H/N |. There exists a
prime q with r/7 < q ≤ r/2 such that q | |N ||G/H|. Let |N |q |G/H|q = q k . If
Qk−1
Qk
|H/N | | i=0 (q k − q i ) with 1 ≤ k ≤ 6, then tr (1) | i=1 (q i − 1). By Lemma 2.1,
the number of prime factors of tr (1) is greater than 6. But the number of primes
Q6
p with p | i=1 (q i − 1) and r/2 < p is less than or equal to 6, a contradiction.
By Lemma 3.4, q k | (r − 1). If gcd(tr (6), r − 1) = 1, then k = 0, contrary to
q | |N ||G/H|. Hence, (i) is true. If gcd(tr (6), r − 1) = p, then k = 1 and q = p. It
follows that (tr (6)/p) | |H/N |. This proves (ii).
Lemma 3.7. Let r ≥ 5. If 1 ≤ N < H ≤ G is a normal series of G, tr (1) | |H/N |,
and H/N is a non-abelian simple group, then H/N ∼
= Ar .
Proof. We consider the following cases:
Case 1. r = 5. In this case, we have |H/N | = 2a 3 · 5 with a ≤ 3. It is clear
that H/N ∼
= A5 .
Case 2. r = 7. In this case, we have |H/N | = 2a 3b 5 · 7 with a ≤ 4 and b ≤ 2.
It is clear that H/N ∼
= A7 .
Case 3. 11 ≤ r ≤ 19. Note that |G| < 1025 for 11 ≤ r ≤ 19. If H/N is not
isomorphic to any alternating group, since tr (1) | |H/N |, by [1, pp. 239–241], H/N
is isomorphic to one of the following groups:
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A. K. Asboei, S. S. S. Amiri, A. Iranmanesh, A. Tehranian
M22 (for r = 11),
Suz (for r = 13),
HS (for r = 11),
M cL (for r = 11),
F i22 (for r = 13),
U6 (2) (for r = 11),
L2 (q) of order ≥ 106 ,
L3 (q) of order ≥ 1012 ,
U3 (q) of order ≥ 1012 ,
L4 (q) of order ≥ 1016 ,
U4 (q) of order ≥ 1016 ,
S4 (q) of order ≥ 1016 ,
G2 (q) of order ≥ 1020 ,
If H/N is isomorphic to one of the six groups on the left side, by |H/N | | |G|,
we have H/N ∼
= M22 and r = 11. So |N |3 |G/H|3 = 3 by |Sr |3 /|M22 |3 = 3.
Since |M22 | - (32 − 3)(32 − 1), we have 3 | 10 by Lemma 3.4, a contradiction.
Suppose H/N is isomorphic to a simple group of Lie type in characteristic p. Let
|H/N |p = pt . If H/N is isomorphic to L4 (q), U4 (q), S4 (q), or G2 (q) of order
≥ 1016 , then pt ≥ 106 by Lemma 4 in [10]. When p ≥ 3, by Lemma 2.3, 106 ≤
pt ≤ p(r−1)/(p−1) ≤ 3(r−1)/2 < 311 , a contradiction. When p = 2, since 219 - |G|,
we have 106 ≤ pt ≤ 218 , a contradiction. If H/N ∼
= U3 (q) (q = pk ), then p 6= 11 by
3k
3
2k
p | |U3 (q)| and 11 - |G|. Thus 11 | p − 1 or 11 | p3k + 1. Since 11 - p2 − 1, we
have exp11 (p) = 5 or 10. Therefore, 5 | k. Thus p3k + 1 has a prime factor ≥ 31
(see Lemma 2 in [11]), contrary to r ≤ 19. Similarly, we derive a contradiction if
H/N ∼
= L2 (q) or L3 (q).
Case 4. 23 ≤ r ≤ 43. Since tr (1) | |H/N |, it is easy to prove that H/N is not
isomorphic to any sporadic simple group. If H/N is isomorphic to a simple group of
Lie type in characteristic 23, we have H/N ∼
= L2 (23),
= L2 (23) or L2 (232 ). If H/N ∼
we have r = 23, since 29 - |L2 (23)|. But 19 - |L2 (23)|, contrary to tr (1) | |H/N |.
If H/N ∼
= L2 (232 ), then r = 43. But 43 - |L2 (232 )|, again contrary to r | |H/N |.
∼
If H/N =3 D4 (pk ) with p 6= 23, then 23 | p8k + p4k + 1 or 23 | p6k − 1. Moreover,
23 | p12k − 1. We have exp23 (p) = 11 or 22 since 23 - p2 − 1. Thus, 11 | k. Then
p132 | |3 D4 (pk )|, contrary to p132 - |G|. If H/N is isomorphic to a simple group
of Lie type in characteristic p except 3 D4 (pk ) with p 6= 23, let |H/N |p = ps . By
examining the orders of simple groups of Lie type, we know that there exists a
positive integer t ≤ s such that 23 | pt + 1 and (pt + 1) | |H/N |, or 23 | pt − 1 and
(pt − 1) | |H/N |. As above, we can prove 11 | t. Thus, s ≥ t ≥ 11. Since p11 - |G|
for r ≤ 43 and p ≥ 5, we have p = 2 or 3. Since 2 and 3 are not primitive roots, we
have (211 − 1) | |H/N | or (311 − 1) | |H/N |. But 211 − 1 and 311 − 1 have a prime
factor > 43, contrary to r ≤ 43.
Case 5. 47 ≤ r ≤ 79. In this case, 47 | |H/N |. It can be proved that H/N is
isomorphic to an alternating group as above.
Case 6. r ≥ 83. Clearly, H/N is not isomorphic to any sporadic simple group
for r ≥ 83. If H/N is isomorphic to a simple group of Lie type in characteristic
p and |H/N |p = pt , then |H/N | < p3t by Lemma 4 in [10]. In particular, if H/N
is not isomorphic to L2 (pt ), then |H/N | < p8t/3 . We first prove p ≤ r/7. If
r/2 < p ≤ r, then we have H/N ∼
= L2 (p). Since |L2 (p)| = p(p2 − 1)/2, the number
of prime factors of tr (1) is not greater than 2, contrary to Lemma 2.1. If r/(s+1) <
p ≤ r/s with s = 2 or 3, then tr (1) < |H/N |/pt < p2t ≤ p2s ≤ p6 ≤ (r/2)6 . But
tr (1) > (r/2)7 by Lemma 2.1, a contradiction. If r/(s+1) < p ≤ r/s with 4 ≤ s ≤ 6,
by Lemma 3.6, we have (2/r)tr (3) < |H/N |/pt < p2t ≤ p2s ≤ p12 ≤ (r/4)12 . By
A characterization of Symmetric group Sr , where r is prime number
19
Lemma 2.1, we have (2/r)tr (3) > (2/r)(r/4)12 t[r/2] (1) > (r/4)12 , a contradiction.
Now we prove that p ≤ r/7 is impossible.
(i) If r ≥ 409 and p ≥ 3, by Lemmas 2.2, 2.3, and 3.6, we have e1.201r <
(2/r)tr (6) < |H/N |/pt < p2t ≤ p2(r−1)/(p−1) < (p2/(p−1) )r ≤ 3r . But e1.201 > 3, a
contradiction.
(ii) For the case where r ≥ 409 and p = 2, if H/N is not isomorphic to L2 (2t ),
we have e1.201r < (2/r)tr (6) < |H/N |/2t < 25t/3 < 25r/3 . But e1.201 > 25/3 , a
contradiction.
Suppose H/N ∼
= L2 (2t ). Since (22t − 1) | |L2 (2t )| and 22t − 1 has a prime factor
q satisfying expq (2) = 2t (see Lemma 2 in [11]), we have 2t + 1 ≤ q ≤ r. Hence,
e1.201r < (r/2)tr (6) ≤ 22t − 1 < 2r , a contradiction.
(iii) If 83 ≤ r ≤ 401 and p ≥ 7, we can deduce e0.775r < 7r/3 as above, a
contradiction.
(iv) If 83 ≤ r ≤ 401 and p ≤ 5, we have 83 | |H/N | by Lemma 3.6. Similar to
the argument used in the case where 23 ≤ r ≤ 43, we can deduce p41 − 1 | |H/N |
or p41 + 1 | |H/N |. But p41 − 1 and p41 + 1 have a prime factor > 401 for p ≤ 5,
contrary to r ≤ 401.
We have proved that H/N ∼
= Ar and G := G/N .
= Ar . Now set H := H/N ∼
On the other hand, we have:
Ar ∼
= HCG (H)/CG (H) ≤ G/CG (H) = NG (H)/CG (H) ≤ Aut(H).
=H∼
∼ G/C (H). Hence Ar ≤ G/K ≤
Let K = {x ∈ G | xN ∈ CG (H)}, then G/K =
G
Aut(Ar ), and hence G/K ∼
= Ar or G/K ∼
= Sr . If G/K ∼
= Ar , then |K| = 2. We
have N ≤ K, and N is a maximal solvable normal subgroup of G, then N = K.
Hence H/N ∼
= Ar = G/N , then |N | = 2. So G has a normal subgroup N of order
2, generated by a central involution z. Therefore G has an element of order 2r.
Now we prove that G does not any element of order 2r, a P
contradiction. At first
we show that r k m2 (Sr ) = m2 (G). We have m2 (Sr ) =
|clSr (xk )| such that
|xk | = 2. Since 2 6= 1, r, the cyclic structure of xk for any k is 1t1 2t2 . . . ltl , where
t1 , t2 , . . . , tl , 1, 2, . . . , l are not equal to r. On the other hand, we have |clSr (xk )| =
r!/1t1 2t2 . . . ltl t1 !t2 ! . . . tl !. Hence m2 (Sr ) = r!h, where h is a real number. Since
m2 (Sr )
r!, then 0 < h < 1. Therefore r k m2 (Sr ). We know that if P and Q
are Sylow r-subgroups of G, then they are conjugate, which implies that CG (P )
and CG (Q) are conjugate. Since 2r ∈ πe (G), we have m2r (G) = φ(2r)nr (G)k =
(r − 1)!k, where k is the number of cyclic subgroups of order 2 in CG (Pr ). Hence
mr (G) | m2r (G). On the other hand, 2r | (1 + m2 (G) + mr (G) + m2r (G)), by (2.1).
Since r | (1+mr (G)) and r | m2 (G), then r | m2r (G). Therefore by (r−1)! | m2r (G)
and r | m2r (G), we can conclude that r! | m2r (G), a contradiction. Hence G/K is
not isomorphic to Ar , and hence G/K ∼
= Sr , then |K| = 1 and G ∼
= Sr . Thus the
proof is completed.
Corollary 3.8. Let G be a finite group. If |G| = |Sr |, where r is a prime number
and |NG (R)| = |NSr (S)|, where R ∈ Sylr (G) and S ∈ Sylr (Sr ), then G ∼
= Sr .
Proof. It follows at once from Theorem 1.
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A. K. Asboei, S. S. S. Amiri, A. Iranmanesh, A. Tehranian
Corollary 3.9. Let G be a finite group. If |NG (P1 )| = |NSr (P2 )| for every prime
p, where P1 ∈ Sylp (G), P2 ∈ Sylp (Sr ) and r is a prime number, then G ∼
= Sr .
Proof. Since |NG (P1 )| = |NSr (P2 )| for every prime p, where P1 ∈ Sylp (G), P2 ∈
Sylp (Sr ), we have |P1 | = |P2 |. Thus, |G|p = |Sr |p for every prime p. Hence,
|G| = |Sr |. It follows that G ∼
= Sr .
4. Proof of the Main Theorem 2
We now prove the theorem 2 stated in the Introduction. Let G be a group such that
nse(G) = nse(Sr ), where r < 5 × 108 and r − 2 are prime numbers and r ∈ π(G).
By Lemma 2.7, we can assume that G is finite. The following lemmas reduce the
problem to a study of groups with the same order with Sr .
Lemma 4.1. If i ∈ πe (Sr ), i 6= 1 and i 6= r, then r k mi (Sr ).
P
Proof. We have mi (Sr ) = |clSr (xk )| such that |xk | = i. Since i 6= 1, r, the cyclic
structure of xk for any k is 1t1 2t2 . . . ltl , where t1 , t2 , . . . , tl , 1, 2, . . . , l are not equal
to r. On the other hand, we have |clSr (xk )| = r!/1t1 2t2 . . . ltl t1 !t2 ! . . . tl !. Hence
mi (Sr ) = r!h, where h is a real number. Since mi (Sr )
r!, then 0 < h < 1.
Therefore r k mi (Sr ).
Lemma 4.2. |Pr | = r.
Proof. At first we prove that if r = 5, then |P5 | = 5. We know that, nse(G) =
nse(S5 ) = {1, 20, 24, 25, 30}. We show that π(G) ⊆ {2, 3, 5}. Since 25 ∈ nse(G),
it follows from (2.1) that 2 ∈ π(G) and m2 = 25. Let 2 6= p ∈ π(G). By (2.1),
we have p ∈ {3, 5, 31}. If p = 31, then by (2.1), m31 = 30. On the other hand,
if 62 ∈ πe (G), then by (2.1), we conclude that m62 = 30 and 62|86, which is a
contradiction. Therefore 62 6∈ πe (G). So P31 acts fixed point freely on the set of
elements of order 2, and |P31 | | m2 , which is a contradiction. Thus π(G) ⊆ {2, 3, 5}.
It is easy to show that, m5 = 24, by (2.1). Also if 3 ∈ πe (G), then m3 = 20. By
(2.1), we conclude that G does not contain any element of order 15, 20 and 25. Also,
we get m4 = 30 and m8 = 24 and G does not contain any element of order 16. Since
2, 5 ∈ π(G), hence we have π(G) = {2, 5} or {2, 3, 5}. Suppose that π(G) = {2, 5}.
Then πe (G) ⊆ {1, 2, 4, 5, 8, 10}. Therefore |G| = 100+20k1 +24k2 +30k3 = 2m ×5n ,
where 0 ≤ k1 + k2 + k3 ≤ 1. Hence 5 | k2 , which implies that k2 = 0, and so
50 + 10k1 + 15k3 = 2m−1 × 5n . Hence 2 | k3 , which implies that k3 = 0. It is easy
to check that the only solution of the equation is (k1 , k2 , k3 , m, n) = (0, 0, 0, 2, 2).
Thus |G| = 22 × 52 . It is clear that πe (G) = {1, 2, 4, 5, 10}, hence exp(P2 ) = 4,
and P2 is cyclic. Therefore n2 = m4 /φ(4) = 30/2 = 15, since every Sylow 2subgroup has one element of order 2, then m2 ≤ 15, which is a contradiction.
Hence π(G) = {2, 3, 5}. Since G has no element of order 15, the group P5 acts
fixed point freely on the set of elements of order 3. Therefore |P5 | is a divisor of
m3 = 20, which implies that |P5 | = 5. Now suppose that r 6= 5, by Lemma 4.1, we
have r2 - mi (G), for any i ∈ πe (G). On the other hand, if r3 ∈ πe (G), then by (2.1)
A characterization of Symmetric group Sr , where r is prime number
21
we have φ(r3 ) | mr3 (G). Thus r2 | mr3 (G), which is a contradiction. Therefore
r3 6∈ πe (G). Hence exp(Pr ) = r or exp(Pr ) = r2 . We claim that exp(Pr ) = r.
Suppose that exp(Pr ) = r2 . Hence there exists an element of order r2 in G such
that φ(r2 ) | mr2 (G). Thus r(r−1) | mr2 (G). And so mr2 (G) = r(r−1)t, where r - t.
If |Pr | = r2 , then Pr will be a cyclic group and we have nr (G) = mr2 (G)/φ(r2 ) =
r(r−1)t/r(r−1) = t. Since mr (G) = (r−1)!, then (r−1)! = (r−1)nr (G) = (r−1)t.
Therefore t = (r − 2)! and mr2 (G) = r(r − 1)(r − 2)! = r!, which is a contradiction.
If |Pr | = rs , where s ≥ 3, then by Lemma 2.6, we have mr2 (G) = r2 l for some
natural number l, which is a contradiction by Lemma 4.1. Thus exp(Pr ) = r. By
Lemma 2.5, |Pr | | (1 + mr (G)) = 1 + (r − 1)!. By [12], |Pr | = r.
Lemma 4.3. π(G) = π(Sr ).
Proof. By Lemma 4.2, we have |Pr | = r. Hence (r − 2)! = mr (G)/φ(r) = nr (G) |
|G|. Thus π((r − 2)!) ⊆ π(G). Now we show that π(Sr ) = π(G). Let p be a
prime number such that p > r. Suppose that pr ∈ πe (G). We have mpr (G) =
φ(pr)nr (G)k, where k is the number of cyclic subgroups of order p in CG (Pr ).
Hence (p − 1)(r − 1)! | mpr . On the other hand, since p is prime and p > r, then
p − 1 > r. Thus (p − 1)(r − 1)! > r!, then mpr > r!, which is a contradiction. Thus
pr 6∈ πe (G). Then Pp acts fixed point freely on the set of elements of order r, and
so |Pp | | (r − 1)!, which is a contradiction. Therefore p 6∈ π(G). By the assumption
r ∈ π(G), hence π(G) = π(Sr ).
Lemma 4.4. G has not any element of order 2r.
Proof. Suppose that G has an element of order 2r. We have
m2r (G) = φ(2r)nr (G)k = (r − 1)!k,
where k is the number of cyclic subgroups of order 2 in CG (Pr ). Hence mr (G) |
m2r (G). On the other hand, 2r | (1 + m2 (G) + mr (G) + m2r (G)), by (2.1). Since
r | (1 + mr (G)) and r | m2 (G) by Lemma 4.1, r | m2r (G). Therefore by (r − 1)! |
m2r (G) and r | m2r (G), we can conclude that r! | m2r (G), a contradiction.
Lemma 4.5. G has not any element of order 3r, 5r, 7r, . . . , pr, where p is the prime
number such that p < r.
Proof. The proof of this lemma is completely similar to Lemma 4.4.
Lemma 4.6. If p = r − 2, then |Pp | = p and np (G) = r!/2p(p − 1).
Proof. Since pr ∈
/ πe (G), then the group Pp acts fixed point freely on the set of
elements of order r, and so |Pp | | mr (G) = (r − 1)!. Thus |Pp | = p. Since Sylow
p-subgroups are cyclic, then np (G) = mp (G)/φ(p) = r!/2p(p − 1).
Lemma 4.7. |G| = |Sr |.
22
A. K. Asboei, S. S. S. Amiri, A. Iranmanesh, A. Tehranian
Proof. We can suppose that |Sr | = 2k2 3k3 5k5 · · · lkl pr, where k2 , k3 , k5 , . . . , kl are
non-negative integers. By Lemma 4.4, the group P2 acts fixed point freely on the
set of elements of order r, and so |P2 | | mr (G) = (r − 1)!. Thus |P2 | | 2k2 . Similarly
by Lemma 4.5, we have |P3 | | 3k3 , . . . , |Pl | | lkl . Therefore |G| | |Sr |. On the other
hand, we know that (r − 2)! = mr (G)/φ(r) = nr (G) and nr (G) | |G| and np (G) =
r!/2p(p−1) | |G|, then the least common multiple of (r −2)! and r!/2p(p−1) divide
the order of G. Therefore r!/2 | |G| and so |G| = |Ar | or |G| = |Sr |. If |G| = |Ar |,
by mr (Sr ) = mr (Ar ) = (r − 1)!, then |NG (R)| = |NAr (S)|, where R ∈ Sylr (G)
and S ∈ Sylr (Ar ), similarly to main Theorem 1, G ∼
= Ar . But we can prove that
nse(G) 6= nse(Ar ). Suppose that nse(G) = nse(Ar ),
Psince nse(G) = nse(Sr ), then
m2 (Sr ) = m2 (Ar ). On the other hand m2 (Sr ) =
|clSr (xi )| such that |xi | = 2,
since cyclic structure 1r−2 2 no exists in Ar , then it is clear that m2 (Sr ) > m2 (Ar ),
a contradiction. Hence |G| = |Sr |.
Now by the main Theorem 1, G ∼
= Sr , and the proof is completed.
Acknowledgment. The authors would like to thank from the referees for the
valuable comments.
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